What is the integral of sin^2(3x)cos^3(3x) ?

The integral of sin^2(3x)cos^3(3x) is 1/3 ((sin^3(3x))/3-(sin^5(3x))/5)+C

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integral of sin^2(3x)cos^3(3x)

\int sin^{2} (3 x) cos^{3} (3 x) d x

\frac{1}{3} (\frac{sin ^{3} ( 3 x )}{3} - \frac{sin ^{5} ( 3 x )}{5}) + C

Solution steps

\int sin^{2} (3 x) cos^{3} (3 x) d x

Apply u-substitution

= \int sin^{2} (u) cos^{3} (u) \frac{1}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sin^{2} (u) cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) sin^{2} (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int v^{2} (1 - v^{2}) d v

Expand

v^{2} (1 - v^{2}) : v^{2} - v^{4}

= \frac{1}{3} \cdot \int v^{2} - v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int v^{2} d v - \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (\frac{v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (\frac{sin ^{3} ( 3 x )}{3} - \frac{sin ^{5} ( 3 x )}{5})

Add a constant to the solution

= \frac{1}{3} (\frac{sin ^{3} ( 3 x )}{3} - \frac{sin ^{5} ( 3 x )}{5}) + C

What is the integral of sin^2(3x)cos^3(3x) ?
The integral of sin^2(3x)cos^3(3x) is 1/3 ((sin^3(3x))/3-(sin^5(3x))/5)+C